[forktheif]

Wikipedia says the USS Ronald Regan has a total crew of 5,680.

How statistically significant is 70 people out of 5,680 suffering serious health problems over 2.5 years? I'm guessing that most people on board were fairly young and in good health.

[skriticos2]

Radiation overdose is happy to cause you all kind of cancer 50 years down the road. Having been overexposed is really not something you want, even if you don't show symptoms so quickly. The psychological pressure on these people alone must be maddening.

Imagine waking up every day for the rest of your life and wondering if you have cancer today? Not good..

Edit: and let's not talk about the increased likelihood of disfigured offspring.

[scotty79]

> Imagine waking up every day for the rest of your life and wondering if you have cancer today? Not good..

30% of us will have it at some point, also you have chance 1 in 1000-2000 that you'll die next year.

[simulate]

> you have chance 1 in 1000-2000 that you'll die next year.

You have a 1000-2000 chance that you'll die when you are no longer an infant but still young. Your chances of dying next year go up as your get older. If you had only 1 chance in 1000 to die each year that would mean that people would live to be 1000 years old on average.

A good estimate of human mortality rates can be calculated from Gompertz law: http://forio.com/simulate/mbean/death-probability-calculator...

[WalterBright]

> If you had only 1 chance in 1000 to die each year that would mean that people would live to be 1000 years old on average.

Actually it means your chances of living to be 1000 would be 37%.

0.999 ^ 1000 = .37

[thaumasiotes]

Your chances of living to be 1000 are approximately 1 in e, 37%, yes. But that has nothing to do with the average age people live to. I don't get 1000 as the average lifespan, though, I get 999. Can someone explain where I'm going wrong with this reasoning?

If there is a 0.001 chance of dying every year, the average lifespan will be 0(0.001) + 1(0.001)(0.999) + 2(0.001)(0.999)^2 + 3(0.001)(0.999)^3 + ... , or \sum_{i=0}^\infty (i(0.001)(0.999)^i). The idea is that to live to the age of 0, you have to die in your first year; to live to be three, you have to live through exactly three years and then die, and so on. Then, \sum_{i=0}^\infty ix^i is x / (1-x)^2 .

So the sum of interest is 0.001 times (0.999) / (0.000001), which is exactly 999.

[magnusjonsson]

You are both correct. See http://en.wikipedia.org/wiki/Exponential_distribution

[scotty79]

I was guessing the age of the parent given this http://ww.reddit.com/r/dataisbeautiful/comments/1t9khp/age_d...

I usually say: I have a chance of 1 in 1000 to die next year. It puts other probabilities in perspective.

[Dylan16807]

>If you had only 1 chance in 1000 to die each year that would mean that people would live to be 1000 years old on average.

You're not old, and you're also not an infant. Your chances are much lower than a naive average.

[lafar6502]

You're being rational again. Stop!